SATHEE: Application of Derivatives 2 Question 29

Application of Derivatives 2 Question 29

####29. Let $f (x) = \begin{array}{cc} x e^{a x}, & x \leq 0 \\ x + a x^{2} - x^{3}, & x > 0 \end{array}$

where, $a$ is a positive constant. Find the interval in which $f^{'} (x)$ is increasing.

(1996, 3M)

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Answer:

Correct Answer: 29. (c, d)

Solution:

NOTE This type is asked in 1983 and repeat after 13 years.

At $x = 0$ , LHL $= lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} x e^{a x} = 0$ and RHL $= lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} (x + a x^{2} - x^{3}) = 0$

Therefore, $LHL = RHL = 0 = f (0)$

So, $f (x)$ is continuous at $x = 0$ .

Also, $f^{'} (x) = \begin{aligned} 1 \cdot e^{a x} + a x e^{a x}, if x < 0 \\ 1 + 2 a x - 3 x^{2}, if x > 0 \end{aligned}$

and $L f^{'} (0) = lim_{x \to 0^{-}} \frac{f (x) - f (0)}{x - 0}$

$= lim_{x \to 0^{-}} \frac{x e^{a x} - 0}{x} = lim_{x \to 0^{-}} e^{a x} = e^{0} = 1$

and

$\begin{aligned} R f^{'} (0) & = lim_{x \to 0^{+}} \frac{f (x) - f (0)}{x + 0} \\ = lim_{x \to 0^{+}} \frac{x + a x^{2} - x^{3} - 0}{x} \\ = lim_{x \to 0^{+}} 1 + a x - x^{2} = 1 \end{aligned}$

Therefore, $L f^{'} (0) = R f^{'} (0) = 1 \Rightarrow f^{'} (0) = 1$ Hence, $f^{'} (x) = \begin{array}{cc} (a x + 1) e^{a x}, & if x < 0 \\ 1, & if x = 0 \\ 1 + 2 a x - 3 x^{2}, & if x > 0 \end{array}$

Now, we can say without solving that, $f^{'} (x)$ is continuous at $x = 0$ and hence on $R$ . We have,

and

$f^{''} (x) = \begin{array}{ll} a e^{a x} + a (a x + 1) e^{a x}, & if x < 0 \\ 2 a - 6 x, & if x > 0 \end{array}$

$L f^{''} (0) = lim_{x \to 0^{-}} \frac{f^{'} (x) - f^{'} (0)}{x - 0}$

$= lim_{x \to 0^{-}} \frac{(a x + 1) e^{a x} - 1}{x}$

$\begin{aligned} = lim_{x \to 0^{-}} a e^{a x} + \frac{e^{a x} - 1}{x} \\ = lim_{x \to 0^{-}} a e^{a x} + a \cdot lim_{x \to 0^{-}} \frac{e^{a x} - 1}{a x} \\ = a e^{0} + a (1) = 2 a \end{aligned}$

and

$\begin{aligned} R f^{''} (0) & = lim_{x \to 0^{+}} \frac{f^{'} (x) - f^{'} (0)}{x + 0} \\ = lim_{x \to 0^{+}} \frac{(1 + 2 a x - 3 x^{2}) - 1}{x} \\ = lim_{x \to 0^{+}} \frac{2 a x - 3 x^{2}}{x} = lim_{x \to 0^{+}} 2 a - 3 x = 2 a \end{aligned}$

Therefore, $L f^{''} (0) = R f^{''} (0) = 2 a$

Hence, $f^{''} (x) = \begin{array}{ll} a (a x + 2) e^{a x}, & if x < 0 \\ 2 a, & if x = 0 \\ 2 a - 6 x, & if x > 0 \end{array}$

Now, for $x < 0, f^{''} (x) > 0$ , if $a x + 2 > 0$

$\Rightarrow$ For $x < 0, f^{''} (x) > 0$ , if $x > - 2 / a$

$\Rightarrow f^{'} (x) > 0$ , if $- \frac{2}{a} < x < 0$

and for $x > 0, f^{''} (x) > 0,$ if $2 a - 6 x > 0$

$\Rightarrow$ for $x > 0, f^{''} (x) > 0$ , if $x < a / 3$

Thus, $f (x)$ increases on $[- 2 / a, 0]$ and on $[0, a / 3]$ .

Hence, $f (x)$ increases on $- \frac{2}{a}, \frac{a}{3}$ .

Application of Derivatives 2 Question 29

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Application of Derivatives 2 Question 29

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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